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Grade: 12th pass

                        

thnx +tan3x=2tan2x find general solution. Please give me ans. Please give me ans

one month ago

Answers : (2)

Vikas TU
12280 Points
							
tan x + tan 2x = tan 3x
Now let us simplify, tan x + tan 2x = tan 3x
tan x + tan 2x – tan 3x = 0
tan x + tan 2x – tan (x + 2x) = 0
On using the formula, tan (A+B) = [tan A + tan B] / [1 – tan A tan B]
Therefore, tan x + tan 2x – [[tan x + tan 2x]/[1- tan x tan 2x]] = 0
(tan x + tan 2x) (1 – 1/(1- tan x tan 2x)) = 0
(tan x + tan 2x) ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0
Now, (tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0
(tan x + tan 2x) = 0 or [– tan x tan 2x] = 0 tan x = tan (-2x) or -tan x tan 2x = 0
tan x = tan (-2x) or 2tan2 x / (1 – tan2 x) = 0
[Using, tan 2x = 2 tan x / 1-tan2 x] x = nπ + (-2x)
or
x = mπ + 0 3x = nπ
or
x = mπ x = nπ/3 or x = mπ
∴ the general solution is x = nπ/3 or mπ,
one month ago
Pradeep Gupta
14 Points
							
tanx + tan 3x = 2tan 2x 
tanx + tan3x = tan2x +tan2x
thnx - tan2x = tan2x - tan3x
tam 2x -tanx =tan 3x - tan2x
tan( 2x-x )( 1 + tan2x tanx ) = tan(3x-2x) ( 1 + tan 3x tan2x)
tanx ( 1+ tan2x tanx ) - tanx ( 1+tan3xtan2x)=0
tanx {1 + tan2xtanx -1-tan3xtan2x}=0
tanx (tan2xtanx- tan3xtan2x)=0
tanx tan2x( tanx -tan3x) =0
If tanx = 0 or tan2x = 0 
x = nπ.           tan2x = nπ
                            x=nπ/2
and tanx - tan3x =0 
tanx = tan3x 
tan3x = tanx 
3x = nπ +x 
2x= nπ
x = nπ /2
 
 
one month ago
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